1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Magnitude of a vector in polar coordinates

  1. Sep 5, 2016 #1
    1. The problem statement, all variables and given/known data
    What is the magnitude of the velocity vector if ##\vec{v} = 4 \hat{r} + 6 \hat{\theta}##

    2. Relevant equations


    3. The attempt at a solution
    I know how do do this in Cartesian coordinates (use the Pythagorean theorem), but not so sure how to do it in polar coordinates.
     
  2. jcsd
  3. Sep 5, 2016 #2

    jambaugh

    User Avatar
    Science Advisor
    Gold Member

    Though the point at which this vector is defined may be given in polar coordinates, the basis expansion [itex]\hat{r},\hat{\theta}[/itex] by which the vector is expressed is still an ortho-normal basis so your vector's magnitude still is the root summed square of components.
     
  4. Sep 5, 2016 #3
    Cool! Now I have a similar question.

    I am given that ##\dot{r} = 4 ~m/s## and ##\dot{\theta} = 2## rad/s, and when a particle is 3 m from the origin, I am asked to find the magnitude of the velocity and the acceleration. This is easy enough.

    From the givens, we know that ##\vec{v} = 4 \hat{r} + 6 \hat{\theta}##. Then we have a magnitude of ##\sqrt{52}##. Then to find the acceleration, we can do either of two things: ##a = \frac{v^2}{r} = \frac{52}{3} = 17.33## m/s. But also, we could use the given information to derive the acceleration vector, which would be ##\vec{a} = -12 \hat{r} + 16 \hat{\theta}##. We find the magnitude of this is 20 m/s^2, not the same as the 17.33 we got earlier. What is going on?
     
  5. Sep 5, 2016 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The acceleration is v2/r in case of uniform circular motion when the velocity is perpendicular to the radius and the acceleration is centripetal only.
     
  6. Sep 5, 2016 #5
    So is the correct answer for the magnitude of the acceleration 17.33 m/s^2?
     
  7. Sep 5, 2016 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    No. Is it uniform circular motion?
     
  8. Sep 5, 2016 #7
    Oops, I meant is the correct magnitude for the acceleration 20 m/s^2? (I looked at the wrong number)
     
  9. Sep 6, 2016 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    That is correct.
     
  10. Sep 6, 2016 #9

    jambaugh

    User Avatar
    Science Advisor
    Gold Member

    An important feature of this problem is that the local rectilinear coordinate system defined by [itex]\hat{r},\hat{\theta}[/itex] is moving. Thus in calculating the acceleration vector you must take into account the time rate of change of the basis vectors.

    In general: [itex] \hat{r} = \hat{i}\cos(\theta) + \hat{j}\sin(\theta)[/itex] so then [itex]\dot{\hat{r}} = \dot{\theta}(-\hat{i}\sin(\theta)+ \hat{j}\cos(\theta)) \equiv \dot{\theta}\hat{\theta}[/itex]. Similarly [itex]\dot{\hat{\theta}} = - \dot{\theta}\hat{r}[/itex]. With these worked out you can take the time derivatives of the position vector to get velocity and acceleration in the local (normalized) coordinate basis:
    Velocity: [itex]\vec{r} = r\hat{r},\quad \vec{v}=\dot{\vec{r}} = \dot{r}\hat{r} + r\dot{\hat{r}}=\dot{r}\hat{r} + r\dot{\theta}\hat{\theta}[/itex]
    Acceleration:
    [itex]\vec{a}=\dot{\vec{v}}=\ddot{r}\hat{r} + \dot{r}\dot{\hat{r}}+ \dot{r}\dot{\theta}\hat{\theta} + r\ddot{\theta}\hat{\theta} + r\dot{\theta}\dot{\hat{\theta}}= [\ddot{r}-r\dot{\theta}^2]\hat{r} +[ 2 \dot{r}\dot{\theta} + r\ddot{\theta}]\hat{\theta} [/itex]

    Knowing your velocity is [itex]4\hat{r} + 6\hat{\theta}[/itex] you know that [itex]\dot{r}=4, r\dot{\theta} = 6[/itex] and given [itex] r = 3meters[/itex] you have [itex]\dot{\theta} = 2[/itex].

    The acceleration vector is then:
    [itex] [\ddot{r}-(3)(2)^2]\hat{r} +[ 2 (4)(2) + (3)\ddot{\theta}]\hat{\theta} = [\ddot{r}-12]\hat{r}+[16+\ddot{\theta}]\hat{\theta}[/itex]
    You have not given enough information to solve the problem. You need the 2nd derivatives of the polar coordinates w.r.t. time. Presumably these are zero??? You haven't stated the full problem as given to you. Was there something stating constant rate of change of theses?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Magnitude of a vector in polar coordinates
Loading...